Given :
Mass of gold ( Au ) is 212 gram.
To Find :
Atoms of gold present in 212 gram of gold.
Solution :
Molecular mass of gold is 197 gram.
So, number of moles of gold in 212 gram is :
Now, we know 1 mole of any element contains 
atoms.
So, number of atoms present in 1.076 moles are :
Hence, this is the required solution.
Answer:
3.76 x 1014 s-1? λ = c/ν = 3.00 x 108 m/s = 7.98 x 10-7 m 3.76 x 1014 s-1.
Explanation:
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Answer:
1. HBr>HCl> H2S >BH3
2.K_a1 very large — H2SO4
K_a1= 1.7 x 10^−2 — H2SO3
K_a1 = 1.7 x 10^−7 — H2S
Explanation:
As one goes down a row in the Periodic Table the properties that determine the acid strength can be observed.
The atoms get larger in radius meaning that in strength, the strength of the bonds get weaker, conversely meaning that the acids get stronger.
For the halogen-containing acids above following the rows and periods, HBr has the strongest bond and is the strongest acid and others follow in this order.
HBr>HCl> H2S >BH3
Acid Dissociation Constant provides us with information known as the ionization constant which comes in handy to measure the acid's strength. The meaning of the proportions are thus, the higher the Ka value, the stronger the acid i.e. it liberates more number of hydrogen ions per mole of acid in solution.
In solution strong acids completely dissociate hence, the value of dissociation constant of strong acids is very high.
Following the cues above on Ka;
K_a1 very large — H2SO4
K_a1= 1.7 x 10^−2 — H2SO3
K_a1 = 1.7 x 10^−7 — H2S
The electrons are unequally shared. The electronegative element receives the electrons from the electropositive one.
